Proposition 5

*Let $X$ be a uniform space. For each Cauchy filter $\mathfrak{F}$ on $X$ there is a unique minimal Cauchy filter $\mathfrak{F}_0$ coarser than $\mathfrak{F}$. If $\mathfrak{B}$ is a base of $\mathfrak{F}$ and $\mathfrak{G}$ is a fundamental system of symmetric entourages of $X$, then the sets $V(M)$ ($M \in \mathfrak{B}, V \in \mathfrak{G}$) form a base of $\mathfrak{F}_0$.

If $M, M'$ are in $\mathfrak{B}$ and $V, V'$ are in $\mathfrak{G}$, then there is a set $M'' \in \mathfrak{B}$ (resp. $V'' \in \mathfrak{G}$) such that $M'' \subset M \cap M'$ (resp. $V'' \subset V \cap V'$); hence $V''(M'') \subset V(M) \cap V'(M')$ and therefore the sets $V(M)$ ($M \in \mathfrak{B}, V \in \mathfrak{G}$) indeed form a base of a filter $\mathfrak{F}_0$ on $X$. Further, if $M$ is $V$-small, then $V(M)$ is $V$-small; hence $\mathfrak{F}_0$ is a Cauchy filter and is clearly coarser than $\mathfrak{F}$. To complete the proof it is enough to show that if $\mathfrak{G}$ is a Cauchy filter coarser than $\mathfrak{F}$, then $\mathfrak{G}$ is finer than $\mathfrak{F}_0$. For each $M \in \mathfrak{B}$ and each $V \in \mathfrak{G}$ there is a set $N \in \mathfrak{G}$ which is $V$-small; since $N \in \mathfrak{F}$, $N$ meets $M$; hence $N \subset V(M)$ and so $V(M) \in \mathfrak{G}$.